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tekilochka [14]
4 years ago
5

Tim is now earning $500 per week after receiving an increase of $45 per week. What was Tim's previous weekly salary

Mathematics
1 answer:
Mars2501 [29]4 years ago
8 0
Now Tim gets $500 after it $45 was added to what he had last week

That means that
"What Tim earned last week" + $45 = $500

Subtract $45 from $500
$500-$45= $455

If you're not sure check your work
If I make $455 now and next week I get $45 more next week
$455+$45=$500
Next week I make $500
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Identify the hypothesis and conclusion of this conditional statement: If two lines intersect at right angles, then the two lines
Sedaia [141]

Answer: b

<u>Step-by-step explanation:</u>

NOTE:

  • Hypothesis: "if" part of the sentence
  • Conclusion: "then" part of the sentence

hypothesis : If <u>two lines intersect at right angles</u>

conclusion: then <u>the two lines are perpendicular</u>

6 0
3 years ago
1. If I get a loan of $30,000 and I promise to pay in 10 equal instalments of $3,000 each semester, how much interest in total d
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Answer:

yes because when you multiply 3000 by 10 that gives you a total of 30000 that you paid off

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4 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
If m∠A = 56°, m∠B = 60°, and c = 8, what is the measure of side length b?
Alekssandra [29.7K]

Using the law of sines:

\frac{b}{sin(B)}=\frac{c}{sin(C)}

Solve for b:

\begin{gathered} b=\frac{c\cdot sin(B)}{sin(C)} \\ so: \\ b=\frac{8\cdot sin(60)}{sin(64)} \\ b\approx7.708 \end{gathered}

Answer:

7.708

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1 year ago
BRAINLIEST GUARANTEED!!! TO WHOEVER SHOWS THEIR WORK
creativ13 [48]

Answer:

B

I

Step-by-step explanation:

A triangle's angles must add up to 180

68+90+x=180

158+x=180

22=x

3 and 2 must add up to 180

70+x=180

angle 2= 110

angle 2 and angle 1 have to add up to 180 as well

110+x=180

andgle 1= 70

3 0
3 years ago
Read 2 more answers
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